Problem: A bag contains $6$ red jelly beans, $2$ green jelly beans, and $4$ blue jelly beans. If we choose a jelly bean, then another jelly bean without putting the first one back in the bag, what is the probability that the first jelly bean will be green and the second will be green as well? Write your answer as a simplified fraction.
Answer: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is picking a green jelly bean and leaving it out. Event B is picking another green jelly bean. Let's take the events one at at time. What is the probability that the first jelly bean chosen will be green? There are $2$ green jelly beans, and $12$ total, so the probability we will pick a green jelly bean is $\dfrac{2} {12}$ After we take out the first jelly bean, we don't put it back in, so there are only $11$ jelly beans left. Also, we've taken out one of the green jelly beans, so there are only $1$ left altogether. So, the probability of picking another green jelly bean after taking out a green jelly bean is $\dfrac{1} {11}$ Therefore, the probability of picking a green jelly bean, then another green jelly bean is $\dfrac{2}{12} \cdot \dfrac{1}{11} = \dfrac{1}{66}$